May-12 (R09) Final by Ali (1-7).p65 - SIA GROUP

Electrical Machines-I (May-2012, R09) JNTU-AnantapurCode No: 9A02308/R09II B.Tech. I Semester Supplementary ExaminationsMay - 2012ELECTRICAL MACHINES-I( Electrical and Electronics Engineering )S.1R09SolutionsTime: 3 Hours Max. Marks: 70Answer any FIVE QuestionsAll Questions carry equal marks- - -1. (a) What is electromechanical energy conversion? (Unit-I, Topic No. 1.1)(b)Develop the block diagram of general electromechanical energy conversion device using energy balanceequation. (Unit-I, Topic No. 1.1)2. Explain the various pitches related to armature winding of a D.C machine. Derive the relation between each for lapand wave windings. (Unit-II, Topic No. 2.3)3. A 1500 kW, 550 V, 16-pole generator runs at 150 r.p.m. What must be the useful flux per pole if there are 2500 lapconnectedconductors and full-load copper losses are 25 kW? Calculate the area of the pole-shoe if the gap fluxdensity has a uniform value of 0.9 T and the no-load terminal voltage, neglecting armature reaction and change inspeed. (Unit-II, Topic No. 2.5)4. What is the significance of critical values of field resistance and speed? Explain how they can be calculated graphicallyfor a D.C generator. (Unit-IV, Topic No. 4.3)5. Two D.C compound generators are operating in parallel with an equalizer-bar connection. Both machines are supplyinga load of 1000 A. The machines have armature resistances of 0.06 Ω and 0.04 Ω, series-field resistances of 0.07 Ω and0.05 Ω and induced e.m.fs of 420 V and 440 V respectively. Calculate,(a) Current in each armature(b) Current in each series-field winding(c) Current in the equalizing-bar(d) Bus-bar voltage. (Unit-V, Topic No. 5.2)6. A 5 kW, 250 V, D.C shunt motor takes no-load armature current of 4 A at rated voltage and runs at 1200 r.p.m. Thearmature circuit resistance is 0.4 Ω and the field resistance is 250 Ω. At rated load and rated voltage, the motor takes26 A and the armature reaction weakens the field flux by 4%. Calculate the full-load speed and the correspondingelectromagnetic torque of the motor. (Unit-VI, Topic No. 6.3)7. The speed of a 50 kW D.C series motor working on 500 V supply is 750 r.p.m at full-load and at 85% efficiency. If theload torque is made 370 N-m and a 2 Ω resistance is connected in series with the machine, calculate the speed atwhich the machine will run. Assume that the magnetic circuit is unsaturated and the total resistance of armature andfield circuits is 0.5 Ω. (Unit-VII, Topic No. 7.1)8. Two shunt machines loaded for the Hopkinson’s test take 15 A at 200 V from the supply. The motor current is 100 Aand the shunt currents are 3 A and 2.5 A. If the resistance of each armature is 0.05 Ω, calculate the efficiency of eachmachine for its particular conditions of loading. (Unit-VIII, Topic No. 8.3.3)WARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.

S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012SOLUTIONS TO MAY-2012, QPQ1. (a) What is electromechanical energy conversion?Answer :May-12, Q1(a)For answer refer Unit-I, Q1.(b) Develop the block diagram of generalelectromechanical energy conversiondevice using energy balance equation.Answer :May-12, Q1(b)For answer refer Unit-I, Q3.Q2. Explain the various pitches related toarmature winding of a D.C machine. Derivethe relation between each for lap and wavewindings.Answer :May-12, Q2Various Pitches Related to Armature Winding(i) Pole PitchFor answer refer Unit-II, Q9, Topic: Pole Pitch.(ii) Coil Span or Coil Pitch (Y S)Coil span or coil pitch is defined as the distancebetween two sides of a coil, calculated in terms of armatureslots or armature conductors.When the coil pitch is equal to the pole pitch thenthat winding is referred as a full-pitched winding. In case ofa full pitched winding coil span is 180° (electrical) and thecoil side lie beneath the opposite poles. Thus, the e.m.fsinduced in these windings may be added.When the coil span is less than the pole pitch, thewinding is referred to as a fractional-pitched winding. Incase of fractional pitched winding the e.m.fs induced in thewindings lying on the two sides of the coil has a phasedifference. Hence, the resultant e.m.f round the coil givenby the vectorial sum of the e.m.fs in the two sides of the coilis comparatively less than the e.m.f induced in case of fullpitchedwinding.(iii) Back Pitch (Y b)For answer refer unit-II, Q9, Topic: Back Pitch.(iv) Front Pitch (Y F)For answer refer Unit-II, Q9, Topic: Front Pitch.(v) Resultant Pitch (Y R)It is defined as a distance measured between thebeginning coil sides of two consecutive coils. The distancemeasured is nothing but the number of conductors presentin between.It is denoted by Y R.For lap winding,Y R= Y B– Y FFor wave winding,Y R= Y B+ Y F(vi) Commutator Pitch (Y C)For answer refer Unit-II, Q9, Topic: Commutator Pitch.Relation between Various Pitches for Lap WindingFor answer refer Unit-II, Q11, Topic: Lap Winding[Note: (i) to (v)]Relation between Various Pitches for Wave WindingFor answer refer Unit-II, Q16, Excluding Topic: DummyCoils.Q3. A 1500 kW, 550 V, 16-pole generator runs at150 r.p.m. What must be the useful flux perpole if there are 2500 lap-connectedconductors and full-load copper losses are 25kW? Calculate the area of the pole-shoe if thegap flux density has a uniform value of 0.9 Tand the no-load terminal voltage, neglectingarmature reaction and change in speed.Answer :May-12, Q3Given that,Lap connected D.C generatorPower, P = 1500 kWSupply voltage, V = 550 VNumber of poles, P = 16Rated speed, N = 150 r.p.mNumber of conductors, Z = 2500Full load copper loss, P cu(f.l)= 25 kWFlux density in the air gap, B = 0.9 TRequired to determine,Flux per pole, φ = ?Area of pole shoe, a = ?No load terminal voltage, E = ?⇒Full load current, I a= VP31500×10=550= 2727.27 AWARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.

Electrical Machines-I (May-2012, R09) JNTU-Anantapur2aFull load copper loss, P cu= I I R a∴ Armature resistance, R a=25×10⇒ R a= 2(2727.27)3Pcu2I a= 3.361117 × 10 –3 ΩNo load terminal voltage,E = V + I aR a= 550 + (2727.27) × 3.361117 × 10 –3 )= 550 + 9.166673= 559.166673 VWe have,Induced e.m.f, E =φZNP60A60AE∴ Flux per pole, φ = ZNP60×16×559.166673⇒ φ=2500×150×16(A = P = 16, Q Lap connected armature)⇒ φ= 0.089466 WbFrom the relation, B = aφflux( φ)∴ Area of pole shoe, a =flux density( B)=0.0894660.9= 0.09940740 m 2= 994.0740 cm 2 .Q4. What is the significance of critical values offield resistance and speed? Explain how theycan be calculated graphically for a D.Cgenerator.Answer :May-12, Q4Critical Field ResistanceFor answer refer Unit-IV, Q19, Topic: Critical FieldResistance.WARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.S.3SignificanceCritical field resistance is the field circuit resistanceabove which the generator fails to excite for any given speed.One of the conditions to be satisfied for self excitation of aD.C generator is that the field circuit resistance must be lessthan the critical field resistance.Critical SpeedFor answer refer Unit-IV, Q22, Topic: Critical Speed.SignificanceFor answer refer Unit-IV, Q22, Topic: Significance.Determination of Critical Field ResistanceFor answer refer Unit-IV, Q19, Topics: Steps to DetermineCritical Field Resistance, Steps to Plot O.C.C Curve.Determination of Critical SpeedFor answer refer Unit-IV, Q20.Q5. Two D.C compound generators are operatingin parallel with an equalizer-bar connection.Both machines are supplying a load of 1000A. The machines have armature resistancesof 0.06 Ω and 0.04 Ω, series-field resistancesof 0.07 Ω and 0.05 Ω and induced e.m.fs of420 V and 440 V respectively. Calculate,(a) Current in each armature(b) Current in each series-field winding(c) Current in the equalizing-bar(d) Bus-bar voltageAnswer :May-12, Q5Given that,For generator G 1,Induced e.m.f, E 1= 420 VArmature resistance, R a1= 0.06 ΩSeries field resistance, R se1= 0.07 ΩFor generator G 2,Induced e.m.f, E 2= 440 VArmature resistance, R a2= 0.04 ΩSeries field resistance, R se2= 0.05 ΩLoad current, I = 1000 ALet, V be the bus bar voltageI a1and I a2be the armature currents of generators ofG 1and G 2respectively.The two compound generators G 1and G 2connectedin parallel which is shown in the figure.

S.4 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012E 2 ,R a2 G 2 G 1E 1 ,R a1+I se2 EqualizingI se1 R se2barR se1VI a2 I a1I eq(a)FigureCurrent in each ArmatureWe have,I = I a1+ I a2= 1000⇒⇒E −V1R a 1420 −V0.06++E −V2R a 2440 −V0.04= 1000= 1000⇒ V = 408 Volts∴ Armature current of generator G 1,I a1==E −V1R a 1420 − 4080.06= 200 AAnd armature current of generator G 2,I a2==E −V2R a 2440 − 4080.04= 800 A(b) Current in each Series WindingThe current through the series field winding of generator,G 1i.e.,⎡I se1= I ⎢⎣ RRse2se1+ Rse2⎡ 0.05 ⎤= 1000 × ⎢ ⎥⎣ 0.07 + 0. 05 ⎦= 416.6666 A⎥⎦⎤–Current in series field winding of generator, G 2i.e.,⎡I se2= I ⎢⎣ Rse1se1+ Rse2WARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.R⎡ 0.07 ⎤= 1000 ⎢ ⎥⎣ 0.07 + 0. 05 ⎦= 583.3333 A(c) Current in the Equalizing-barCurrent in the equalizing bar, I eqfrom generator side (2),we have,I a2= I se2+ I eq⇒ I eq= I a2– I se2= 800 – 583.3333= 216.6667 A∴ Current in the equalizing bar,I eq= 216.6667 A flows from generator G 2togenerator G 1(d) Bus-bar VoltageWe have,V = E 1– I a1R a1– I se1R se1= 420 – 200 × 0.06 –416.6666 × 0.07= 378.8333 VResult(a) Current in each armature,I a1= 200 AI a2= 800 A(b) Current in each series field winding,I se1= 416.6666 AI se2= 583.3333 A(c) Current in equalizing bar,I eq= 216.6667 A(d) Bus bar voltage,V = 378.8333 VQ6. A 5 kW, 250 V, D.C shunt motor takes no-loadarmature current of 4 A at rated voltage andruns at 1200 r.p.m. The armature circuitresistance is 0.4 Ω and the field resistance is250 Ω. At rated load and rated voltage, themotor takes 26 A and the armature reactionweakens the field flux by 4%. Calculate thefull-load speed and the correspondingelectromagnetic torque of the motor.May-12, Q6⎥⎦⎤

Electrical Machines-I (May-2012, R09) JNTU-AnantapurAnswer :Given that,D.C shunt motorVoltage, V = 250 VoltsPower output, P = 5 kWNo.-load armature current, I a0= 4 ANo.-load speed, N o= 1200 r.p.mArmature resistance, R a= 0.4 ΩShunt field resistance, R sh= 250 ΩFull load current, I L= 26 AReduction in flux = 4 %Required to determine,(i) Full-load speed, N F.L= ?(ii) Torque, τ = ?No load back e.m.f, EShunt current,I sh=boVR sh= V – I aoR a= 250 – 4 × 0.4= 248.4 Volts250= 250= 1 AFull-load armature current,I af= I L– I sh= 26 – 1= 25 AFull-load back e.m.f, E = V – I a fb fR a= 250 – 25 × 0.4= 240 VoltsReduction in flux = 4 % (given)⇒ φ f= 0.96 φ oWhere,φ fis the full load fluxφ ois the no-load flux⇒We have,∴φφofN ∝N F . L=No=10.96E bφEEb fbo×φφof⇒ N F.L= N o×WARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.EEb fbo×φφ240 1= 1200 × ×248.4 0. 96= 1207.7295 r.p.mN F L∴ Full load speed, . = 1207.7295 r.p.mWe also have,Torque, τ = 9.55 ×= 9.55 ×EbN= 47.445 N-mfIafF.Lo240× 251207.7295∴ Torque = 47.445 N−mfS.5Q7. The speed of a 50 kW D.C series motorworking on 500 V supply is 750 r.p.m at fullloadand at 85% efficiency. If the load torqueis made 370 N-m and a 2 Ω resistance isconnected in series with the machine,calculate the speed at which the machinewill run. Assume that the magnetic circuit isunsaturated and the total resistance ofarmature and field circuits is 0.5 Ω.Answer :May-12, Q7Given that,D.C series motor,Output power, P = 50 kWInput voltage, V = 500 VFull-load speed, N F.L= 750 r.p.metaFull-load efficiency, η F. L= 85 % = 0.85New load torque, τ 2= 370 N.mExternal resistance, R ext= 2 ΩArmature and field resistance, R = 0.5 ΩAlso given that the magnetic circuit is unsaturated.Required to determine,New speed, N 2= ?At full load condition,P2 πN F .LFull load torque, τ F.L=60=350×102π×750= 636.6198 N.m60

S.6 Spectrum ALL-IN-ONE Journal for Engineering Students, 2012⇒Full load efficiency,η F.L===OutputInputPVI LF . LPVI aF . LArmature current,I aF . LP=VηF.L50×10=500×0.85[Q For series motor, I L= I a= I se]3= 117.647 ABack e.m.f with no external resistance added,E.= V – (b F LI aF . L× R)= 500 – (117.647 × 0.5)= 441.1765 VoltsAt new load conditions,New load torque, τ 2= 370 N-mExternal resistance, R ext= 2 ΩWe have, τ ∝ φ I a2⇒ τ ∝ I a[Q φ −~ I a]∴⇒2F . LI aF. L=τ22 I a2τ2I a 2=⇒ I a =2ττ2F.Lττ= I aF . L2F.L×2I aF .L× Iττ2a F .L2F.LSpeed of D.C motor is given by the relation,N ∝⇒ N =⇒N2N F . L=E bφEIEbaEb2b F . L⇒ N 2= N F.L××EE[Q φ −~ I a]IaF. LIb2b F . La2×IaF. L275.77625 117.647⇒ N 2= 750 ××441.1765 89.6895= 614.9573 r.p.m∴ New speed= 614.9573 r.p.m.Q8. Two shunt machines loaded for theHopkinson’s test take 15 A at 200 V from thesupply. The motor current is 100 A and theshunt currents are 3 A and 2.5 A. If theresistance of each armature is 0.05 Ω,calculate the efficiency of each machine forits particular conditions of loading.Answer :May-12, Q8Given that,Two shunt machines for Hopkinson’s test,Supply voltage, V = 200 VLine current, I L= 15 AMotor input current, I m= 100 AShunt field current of generator, I sh g= 3 AShunt field current of motor, I sh = 2.5 AmArmature resistance, R a= 0.05 ΩRequired to determine,Efficiency of each machine,η g= ?η m= ?Figure (1) shows the motor generator set drawn fromthe given data.Ia2370= 117.647 ×636.6198= 89.6895 ABack e.m.f when an external resistance of 2 Ω is addedin series with the machine,E b 2= V – [ I a × (R + R2ext)]VI LI shmMI MIa mI gI agI gGI shg= 500 – [89.6895 × (0.5 + 2)]= 275.77625 VoltsFigure (1)WARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.

Electrical Machines-I (May-2012, R09) JNTU-Anantapur∴Motor armature current,I a m= I m– I sh m= 100 – 2.5= 97.5 AGenerator output current,I g= I M– I L= 100 – 15 = 85 AGenerator armature current,I a g= I g += 85 + 3= 88 AMotor copper loss,P cu(m)I sh g= Armature copper loss + Shunt field loss=2I a m× R a+ V I sh m= [(97.5) 2 × 0.05] + [200 × 2.5]= 475.3125 + 500= 975.3125 WGenerator copper loss,P cu(g)=2I a g× R a+ V I sh g= [(88) 2 × 0.05] + [200 × 3]= 387.2 + 600= 987.2 WCopper losses for the M-G set,Motor output = Motor input – Total losses= (V × I m) – 1494.05625= (200 × 100) – 1494.05625= 20000 – 1494.05625= 18505.94375 W∴ Motor efficiency, η m=Motor outputMotor input× 100S.718505.94375=× 10020000= 0.925297 × 100= 92.5297 %GeneratorTotal losses = Generator copper loss + Stray loss= 987.2 + 518.74375= 1505.94375Generator output = V I g= 200 × 85= 17000 WGenerator input = Generator output + Total losses= 17000 + 1505.94375= 18505.94375 W∴P cu= P cu(m)+ P cu(g)= 975. 3125 + 987.2= 1962.5125 WInput power to the set = V × I L= 200 × 15 = 3000 WStray losses for the M-G set,P s= Input – Copper losses= 3000 – 1962.5125= 1037.4875 WGenerator efficiency, η g==Output × 100Input1700018505.94375= 0.918623 × 100= 91.8623 %× 100Motor∴ Stray losses for each machine =1037.48752= 518.74375 WTotal losses = Motor copper losses + Stray losses= 975.3125 + 518.74375= 1494.05625 WWARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.